# 10 - 5 Mass of plate =(4R2 mass of hole =(R/2)2 = R2/4 xCM...

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x CM = 5. Mass of plate = σ (4R 2 ), mass of hole = – σ π (R/2) 2 = – σ π R 2 /4. x CM = [4R 2 (0) – ( π R 2 /4)(R/2)]/(4 – π /4) σ R 2 = –0.122 R = y CM 6. x CM = [0 – ρ 4 π r 3 d/3]/(R 3 – r 3 ) 4 π ρ = –r 3 d/(R 3 – r 3 ) 15. m collides with the box. p :–mu = 4mv 1 , so v 1 = –u/4. In the time t 1 = L/u it took m to collide with box, 2m moves L/2. The relative speed of the box and 2m is (u/2 + u/4) = 3u/4. The box moves for a time (L/2) (3u/4) = 2L/3u, so its displacement Δ x = (2L/3u) (–u/4) = –L/6. 24. v CM = (7i + 4 j   )/7, and r 2 = 0 = r 1 + v CM Δ t, so r 1 = –3i – 1.71 j   m 28. (a) v CM = v/21, K rel = 0.5(1 u )(20v/21) 2 + 0.5(20 u)(v/21) 2 Thus E = (10v 2 /21)(1.66 × 10 –27 kg) K i = 1/2 (1 u)v 2 = 1/2 (21E/10) = 8.4 × 10 –19 J; (b) v CM = 20v/21, K rel = 0.5(1 u)(20v/21) 2 + 0.5(20u ) (v/21) 2 . Thus E = (10v 2 /21) (1.66 × 10 –27 kg) K i = 1/2(20u)v 2 = 21E = 1.68 × 10 –17 J 31. v CM = [4 ×  5 j   + 0]/8 = 2.5 j   m/s. So, v 1 = 1.5 j   m/s, and v 2 = –2.5 j m/s. (a)K Rel = 1/2(3)(2.5) 2 + 1/2(5)(1.5) 2 = 15 J; (b) K CM = 1/2(8)(2.5) 2 = 25 J. 32. (a) T = v ex dm/dt = (2.6 × 10 3 ) (1.5 × 10 4 ) = 3.9 × 10 7 N;

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10 - 5 Mass of plate =(4R2 mass of hole =(R/2)2 = R2/4 xCM...

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